What is the standard voltage generated by the corrosion reaction on the iron hull of a ship? — Electrochemistry Chemistry Question

💡 Solution & Explanation

**Step 1: Identify the half-reactions in iron hull corrosion.** **Cathode (reduction):** Oxygen reduction in the presence of water: $$\ce{O2(g) + 4H+(aq) + 4e- -> 2H2O(l)}, \quad E^\circ = +1.229\ \text{V}$$ **Anode (oxidation):** Dissolution of iron: $$\ce{Fe(s) -> Fe^{2+}(aq) + 2e-}, \quad E^\circ_{ox} = +0.447\ \text{V}$$ The standard reduction potential of iron is $E^\circ(\text{Fe}^{2+}/\text{Fe}) = -0.447\ \text{V}$. **Step 2: Calculate the standard cell potential.** $$E^\circ_{\text{cell}} = E^\circ_{\text{cathode}} - E^\circ_{\text{anode(red)}}$$ $$E^\circ_{\text{cell}} = 1.229 - (-0.447) = 1.229 + 0.447 = \mathbf{+1.676\ \text{V}}$$ The positive value confirms the corrosion reaction is **thermodynamically spontaneous**. **Step 3: Check the options.** - A: $-1.676\ \text{V}$ — incorrect sign - B: $-0.782\ \text{V}$ — incorrect - C: $+1.782\ \text{V}$ — incorrect - D: $+1.676\ \text{V}$ ✓ $$\boxed{\text{Answer: D — } E^\circ_{\text{cell}} = +1.676\ \text{V}}$$